Sunday, 14 April 2019

calculus - Proof that Γ(1)=γ?



I know that Γ(1)=γ, but how does one prove this?
Starting from the basics, we have that:




Γ(x)=0ettx1dt



How do we differentiate this? How do we then find that



Γ(1)=0etlog(t)dt



and how would one solve this integral?


Answer



Taken from this answer:




By the recursive relation Γ(x+1)=xΓ(x), we get
log(Γ(x))=log(Γ(n+x))log(x)log(x+1)log(x+2)log(x+n1)


Differentiating (1) with respect to x, evaluating at x=1, and letting n yields
Γ(1)Γ(1)=log(n)+O(1n)1112131nγ

A discussion of why ddxlog(Γ(x))=log(x)+O(1x), and a proof of the log-convexity of Γ(x), is given in the answer cited above. It is easy to accept since log(Γ(n+1))=log(Γ(n))+log(n).






Another Approach
0log(t)etdt=limnn0log(t)(1tn)ndt=limnn10(log(t)+log(n))(1t)ndt=limn(nn+1log(n)+n10log(1t)tndt)=limn(nn+1log(n)nn+1Hn+1)=γ


Explanation:
(3a): Monotone Convergence and et=limn(1tn)n
(3b): substitute tnt
(3c): substitute t1t
(3d): integrate by parts 10log(1t)tndt=1n+1101tn+11tdt
(3e): γ=limnHnlog(n)


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